Question

What is the partial fraction expansion of the proper function X(z)=\(\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\)?

(a) \(\frac{2z}{z-1}-\frac{z}{z+0.5}\)

(b) \(\frac{2z}{z-1}+\frac{z}{z-0.5}\)

(c) \(\frac{2z}{z-1}+\frac{z}{z+0.5}\)

(d) \(\frac{2z}{z-1}-\frac{z}{z-0.5}\)

This question was addressed to me in semester exam.

My question comes from Inversion of Z Transform in section Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

Correct answer is (d) \(\frac{2z}{z-1}-\frac{z}{z-0.5}\)

Explanation: First we eliminate the negative powers of z by multiplying both numerator and denominator by z^2.

Thus we obtain X(z)=\(\frac{z^2}{z^2-1.5z+0.5}\)

The poles of X(z) are p1=1 and p2=0.5. Consequently, the expansion will be

\(\frac{X(z)}{z} = \frac{z}{(z-1)(z-0.5)} = \frac{2}{(z-1)} – \frac{1}{(z-0.5)}\)  (obtained by applying partial fractions)

=>X(z)=\(\frac{2z}{(z-1)}-\frac{z}{(z-0.5)}\).